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Chapter 1: Problem 41

A sample of calcium carbonate \(\left(\mathrm{CaCO}_{3}\right)\) has the following percentage composition : \(\mathrm{Ca}=40 \%\), \(\mathrm{C}=12 \%, \mathrm{O}=48 \%\). If the law of constant proportions is true, then the weight of calcium in \(4 \mathrm{~g}\) of a sample of calcium carbonate obtained from another source will be : (a) \(0.016 \mathrm{~g}\) (b) \(0.16 \mathrm{~g}\) (c) \(1.6 \mathrm{~g}\) (d) \(16 \mathrm{~g}\)

Short Answer

Expert verified
The weight of calcium in a 4 g sample of calcium carbonate is 1.6 g.

Step by step solution

01

Understand the Law of Constant Proportions

The Law of Constant Proportions, also known as the Law of Definite Proportions, states that in a chemical substance the elements are always combined in the same proportion by mass, regardless of the source or how the compound was prepared. This means that any sample of calcium carbonate will have the same percentage composition of calcium, carbon, and oxygen.
02

Calculate the Weight of Calcium in the Sample

Knowing that the percentage composition of calcium in calcium carbonate is 40%, we simply need to calculate 40% of the given weight of the sample to find out how much calcium is in a 4 g sample. This is done by multiplying the sample's weight by the percentage of calcium: \[ \text{Weight of calcium} = 4 \text{ g} \times \frac{40}{100} = 1.6 \text{ g} \]
03

Choose the Correct Answer

From the calculation in the previous step, we find that the weight of calcium in a 4 g sample of calcium carbonate is 1.6 g. This corresponds to option (c).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Composition
Chemical composition refers to the identity and quantity of the elements that make up a chemical compound. For example, water (H2O) consists of hydrogen and oxygen in a 2:1 ratio by atom count. Understanding chemical composition is vital for many scientific inquiries, including chemistry, biology, and environmental science.

Using the law of constant proportions, we know that no matter where we obtain a sample of calcium carbonate (CaCO3), it will always be composed of calcium (Ca), carbon (C), and oxygen (O) in the same ratio by mass. In this case, if a sample has the percentages Ca=40%, C=12%, and O=48%, these reflect how the mass of the compound is distributed among its constituent elements.
Percentage Composition
The percentage composition of a compound represents the percent by mass of each element in the compound. This concept is crucial when determining the formula of a compound based on the masses of its elements and vice versa. For example, in a sample of calcium carbonate (CaCO3), if the percentage composition for calcium (Ca) is 40%, it means that in a 100 g sample of calcium carbonate, calcium would account for 40 g.

When a question asks for the mass of calcium in a different sample size, such as 4 g of calcium carbonate, applying the percentage composition is straightforward. You would calculate 40% of 4 g to determine the mass of calcium in this new sample, illustrating how the percentage composition helps in scaling the mass of elements in different sample sizes of the same compound.
Stoichiometry
Stoichiometry deals with the quantitative relationships of the elements and compounds involved in chemical reactions. It is based on the laws of conservation of mass and constant proportions. In stoichiometry, understanding the mole concept, molar mass, and Avogadro's number proves to be indispensable. For instance, reacting elements in a specific ratio by moles will yield products in a predictable ratio, adhering to the established reaction equation.

In the context of our calcium carbonate example, stoichiometry would involve calculating the number of moles of calcium based on its molar mass and using the percentage composition to find out how much of it would be present in any given amount of the compound. Through stoichiometry, we can predict the outcome of reactions involving calcium carbonate, including how much product will form from a certain amount of reactant, emphasizing the practicality of stoichiometric calculations in real-world applications.

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Most popular questions from this chapter

Which of the following setups is correct to calculate the weight (in \(\mathrm{g}\) ) of \(\mathrm{KClO}_{3}\) produced from the reaction of \(0.150\) moles of \(\mathrm{Cl}_{2}\) ? $$ 3 \mathrm{Cl}_{2}+6 \mathrm{KOH} \longrightarrow 5 \mathrm{KCl}+\mathrm{KClO}_{3}+3 \mathrm{H}_{2} \mathrm{O} $$ (a) \(0.150\) moles \(\mathrm{Cl}_{2} \times 1\) mole \(\mathrm{KClO}_{3} / 3\) moles \(\mathrm{Cl}_{2} \times 122.5 \mathrm{~g} / 1\) mole \(\mathrm{KClO}_{3}\) (b) \(0.150\) moles \(\mathrm{Cl}_{2} \times 1\) mole \(\mathrm{KClO}_{3} / 3\) moles \(\mathrm{Cl}_{2} \times 1\) mole \(\mathrm{KClO}_{3} / 122.5 \mathrm{~g}\) (c) \(0.150\) moles \(\mathrm{Cl}_{2} \times 3\) moles \(\mathrm{Cl}_{2} / 1\) mole \(\mathrm{KClO}_{3} \times 122.5 \mathrm{~g} / 1\) mole \(\mathrm{KClO}_{3}\) (d) \(0.150\) moles \(\mathrm{Cl}_{2} \times 3\) moles \(\mathrm{Cl}_{2} / 1\) mole \(\mathrm{KClO}_{3} \times 1\) mole \(\mathrm{KClO}_{3} / 122.5 \mathrm{~g}\)

The formula of an acid is \(\mathrm{H} X \mathrm{O}_{2}\). The mass of \(0.0242\) moles of the acid is \(1.657 \mathrm{~g}\). What is the atomic weight of \(X ?\) (a) \(35.5\) (b) \(28.1\) (c) 128 (d) \(19.0\)

Total no. of electrons present in \(48 \mathrm{~g} \mathrm{Mg}^{2+}\) are: (a) \(24 N_{A}\) (b) \(2 N_{A}\) (c) \(20 \mathrm{~N}_{A}\) (d) none of these

A \(25.0 \mathrm{~mm} \times 40: 0 \mathrm{~mm}\) piece of gold foil is \(0.25 \mathrm{~mm}\) thick. The density of gold is \(19.32 \mathrm{~g} / \mathrm{cm}^{3}\). How many gold atoms are in the sheet? (Atomic weight : \(\mathrm{Au}=197.0\) ) (a) \(7.7 \times 10^{23}\) (b) \(1.5 \times 10^{23}\) (c) \(4.3 \times 10^{21}\) (d) \(1.47 \times 10^{22}\)

The oxidation number of oxygen in \(\mathrm{KO}_{3}, \mathrm{Na}_{2} \mathrm{O}_{2}\) is : (a) 3,2 (b) 1,0 (c) 0,1 (d) \(-0.33,-1\)
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